# Calculate Drawdown in Python Language Assignment Solution

July 04, 2024
Professor Daniel
Python
l Professor Daniel Mitchell is an esteemed faculty member with a Master's degree in Software Engineering from a prestigious university in Canada. He has successfully completed over 700 assignments focusing on Python file operations. Professor Mitchell's expertise lies in file system design, secure file handling practices, and integrating file operations with database systems. His insights into binary file processing and path management make him a valuable resource for students seeking comprehensive Python file handling solutions.
Key Topics
• Instructions
• Objective
• Requirements and Specifications
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## Instructions

### Objective

Write a Python assignment program to calculate drawdown in the Python language. This program aims to determine the drawdown, which is a measure of the decline from a peak value to a trough value in a time series. By writing a Python assignment that calculates drawdown, you'll be able to analyze the magnitude of losses within the given data. This assignment provides an opportunity to showcase your understanding of Python programming concepts and your ability to manipulate financial data for analytical purposes.

## Requirements and Specifications

Source Code

```#!/bin/python3 import math import os import random import re import sys # # Complete the 'drawdowncalc' function below. # # The function is expected to return a FLOAT. # The function accepts following parameters: # 1. FLOAT_ARRAY rets # 2. INTEGER N # def findmax(all_amounts): maxdrawdowncalc = None maxdrawdowncalc_int = None maxdrawdowncalc_index = -1 for k in range(len(all_amounts)): amounts = all_amounts[k] l = len(amounts) for i in range(l-1): for j in range(i+1,l): drawdown = (amounts[i]-amounts[j])/amounts[i] if drawdown > 0: if (maxdrawdowncalc is None) or (drawdown > maxdrawdowncalc): maxdrawdowncalc = drawdown maxdrawdowncalc_int = [i,j] maxdrawdowncalc_index = k new_amounts = [] for k in range(len(all_amounts)): if k != maxdrawdowncalc_index: new_amounts.append(all_amounts[k]) else: new_amounts.append(all_amounts[k][:maxdrawdowncalc_int[0]+1]) new_amounts.append(all_amounts[k][maxdrawdowncalc_int[1]:]) return maxdrawdowncalc, new_amounts def drawdowncalc(rets, N): curr = 1 amounts = [curr] for i in range(len(rets)): curr *= (1 + rets[i]) amounts.append(curr) result = None new_amounts = [amounts] for i in range(N): result, new_amounts = findmax(new_amounts) if result is None: return i return -round(result, 4) if __name__ == '__main__': l = int(input()) rets = [] for _ in range(l): rets.append(float(input())) n = int(input()) print(drawdowncalc(rets, n)) # test cases # print(drawdowncalc([0.01, -0.01, 0.004, -0.02, 0.01], 1)) # print(drawdowncalc([0.01, -0.01, -0.01, 0.05, -0.04, 0.01], 2)) # print(drawdowncalc([0.01, -0.04, 0.05, -0.01, -0.01, 0.01], 3)) # print(drawdowncalc([0.01, 0.01, 0.01, -0.01, 0.01, 0.01], 1)) # print(drawdowncalc([0.01, 0.02, 0.03, -0.01, -0.02, 0.04, 0.05, -0.02, -0.02, 0.03], 2)) ```

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